Sokhotski–Plemelj formulae

Remarks

Proof idea: write $\int_\Gamma \frac{\varphi(\tau)}{\tau - z} \, d\tau = \int_\Gamma \frac{\varphi(\tau) - \varphi(t)}{\tau - z} \, d\tau + \varphi(t) \int_\Gamma \frac{d\tau}{\tau - z}$. The last integral is $2\pi i$ for $z \in D^+$ and $0$ for $z \in D^-$. The Hölder condition makes the first integral continuous up to $\Gamma$ from both sides, with value $\mathrm{p.v.} \int_\Gamma \frac{\varphi(\tau) - \varphi(t)}{\tau - t} \, d\tau$ at $t$. Finally, $\mathrm{p.v.} \int_\Gamma \frac{d\tau}{\tau - t} = \pi i$, half of $2\pi i$, because the excised small arc around $t$ leaves half a turn.

Example: on the unit circle with $\varphi(\tau) = \tau^{-1}$, one has $\Phi = 0$ on $D^+$ and $\Phi(z) = -1/z$ on $D^-$. So $\Phi^+(t) - \Phi^-(t) = t^{-1} = \varphi(t)$, and the principal value integral equals $\pi i (\Phi^+(t) + \Phi^-(t)) = -\pi i \, t^{-1}$.

Continuity of $\varphi$ alone is not enough: there are continuous densities whose principal value integral fails to exist at some points. The formulae also hold on a smooth open arc, at every point other than the endpoints. There $\Phi^+$ is the limit from the left of the arc and $\Phi^-$ the limit from the right. For the real line they reduce to the Sokhotski–Plemelj theorem.

Going deeper

For a student

The formulae are the bridge between boundary value problems for holomorphic functions and singular integral equations. The Riemann–Hilbert problem $\Phi^+ = G \Phi^- + g$ on $\Gamma$ is solved by writing the unknowns as Cauchy-type integrals. Conversely, substituting $\Phi^\pm$ into the formulae turns the singular integral equation $a \varphi + b S \varphi = f$, with $S \varphi = \frac{1}{\pi i} \mathrm{p.v.} \int_\Gamma \frac{\varphi(\tau)}{\tau - t} d\tau$, into the boundary condition $(a + b) \Phi^+ - (a - b) \Phi^- = f$. This reduction is the basis of the theory of the index of a singular integral operator.